Two point charges of 2 C each are separated from each other by a distance of 1 m in water (K = 80). The force between the charges is

$4.5 × 10^8 N$

The correct answer is Option (2) → $4.5 × 10^8 N$

Coulomb's law in a medium: $F = \frac{1}{4\pi \epsilon_0 K} \frac{q_1 q_2}{r^2}$

Given: $q_1 = q_2 = 2 \, C$, $r = 1 \, m$, $K = 80$, $\epsilon_0 = 8.854 \times 10^{-12} \, F/m$

$F = \frac{1}{4\pi (8.854 \times 10^{-12} \times 80)} \frac{2 \cdot 2}{1^2}$

$F = \frac{4}{4\pi \cdot 7.0832 \times 10^{-10}}$

$F \approx \frac{4}{8.885 \times 10^{-9}} \approx 4.5 \times 10^8 \, N$

Force $F \approx 4.5 \times 10^8 N$